Reply To: If $$tanA = \frac{sinB}{1-cosB}$$ find tan2A
To find \( \tan(2A) \) given that \( \tan(A) = \frac{\sin(B)}{1 – \cos(B)} \), we’ll first express \( \tan(A) \) in a more familiar trigonometric identity and then use the double angle formula for \( \tan \).
Given:
\[ \tan(A) = \frac{\sin(B)}{1 – \cos(B)} \]
Recall the trigonometric identity:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin(\theta)}{1 + \cos(\theta)} \]
However, our expression looks similar but not exactly the same. Notice that for \( \frac{\sin(B)}{1 – \cos(B)} \), if we consider \( \theta = 2B \), we get:
\[ \tan\left(\frac{2B}{2}\right) = \frac{\sin(2B)}{1 + \cos(2B)} \]
But our denominator is \(1 – \cos(B)\), not \(1 + \cos(B)\). To relate this to the given equation, let’s use the identity:
\[ \tan(A) = \frac{1 – \cos(\theta)}{\sin(\theta)} \]
This is a form of the half-angle identity. However, to directly find \( \tan(2A) \) from \( \tan(A) = \frac{\sin(B)}{1 – \cos(B)} \), let’s directly apply the double angle formula for tangent, knowing that \( \tan(A) \) is given. The double angle formula is:
\[ \tan(2A) = \frac{2\tan(A)}{1 – \tan^2(A)} \]
Substituting \( \tan(A) = \frac{\sin(B)}{1 – \cos(B)} \) into the formula:
\[ \tan(2A) = \frac{2\left(\frac{\sin(B)}{1 – \cos(B)}\right)}{1 – \left(\frac{\sin(B)}{1 – \cos(B)}\right)^2} \]
Let’s simplify this expression.
After simplifying the expression, we find that:
\[ \tan(2A) = -\tan(B) \]
This result shows that \( \tan(2A) \) is directly related to \( -\tan(B) \), given the initial relationship between \( \tan(A) \) and \( \sin(B) \) and \( \cos(B) \).